Metrics H andbehavioursas sk1 2on loopgroups · 2017-02-08 ·...

Journal of Functional Analysis 213 (2004) 440–465

Metrics Hs and behaviours as sk1=2 onloop groups

Shizan Fang

Laboratoire de Topologie, Universite de Bourgogne, 9 avenu Alain Savary, BP 47870, Dijon Cedex,

France and Department of Mathematics, Beijing Normal University, Beijing 100875, China

Received 10 June 2003; accepted 23 July 2003

Communicated by Paul Malliavin

Abstract

For all metricHs with s41=2; the heat measures on free loop groups have been constructedby P. Malliavin. The geometric stochastic analysis is well developed. The metricH1=2 is critical

and involves naturally in Mathematical Physics. The Ricci tensors Rics play the fundamental

role in these situations. In this note, we shall study the behaviour of Rics as sk1=2:r 2003 Elsevier Inc. All rights reserved.

0. Introduction

Let G be a semi-simple compact Lie group and G its Lie algebra, endowed with an

AdG invariant metric /;SG: Denote by D ¼ � d2

dy2the Laplace operator on the circle

S1: For s40; define on CNðS1;GÞ the following metric:

jhj2s ¼Z

S1

/ð1þ DÞsh; hSG

dy2p

:

Writing hACNðS1;GÞ in Fourier series,

hðyÞ ¼XnAZ

hðnÞeffiffiffiffiffi�1

pny where hðnÞ ¼

ZS1

hðyÞe�ffiffiffiffiffi�1

pny dy2p

; ð0:1Þ

ARTICLE IN PRESS

E-mail address: [email protected].

0022-1236/$ - see front matter r 2003 Elsevier Inc. All rights reserved.

doi:10.1016/j.jfa.2003.07.006

then

jhj2s ¼XnAZ

ð1þ n2ÞsjhðnÞj2G: ð0:2Þ

Let

HsðGÞ ¼ fhAL2ðS1;GÞ; jhjsoþNg: ð0:3Þ

By Sobolev embedding theorem, HsðGÞCCðS1;GÞ for s41=2: In this case, there

is a Gaussian measure ms on CðS1;GÞ such that ðCðS1;GÞ;HsðGÞ; msÞ becomes anabstract Wiener space. Let fun; nX1g be an orthonormal basis of HsðRÞ;fe1;y; edg an orthonormal basis of G: Define

en;aðyÞ ¼ unðyÞea; nX1; a ¼ 1;y; d: ð0:4Þ

Then fen;ag is an orthonormal basis of HsðGÞ: In what follows, we shall work with

this type of basis. Let GðsÞðy1; y2Þ be the Green function on S1 associated to the

differential operator ð1þ DÞs; that is the solution of

ð1þ DÞsGðsÞðy1; Þ ¼ dy1 ; ð0:5Þ

where dy1 is the Dirac mass at y1: Relation (0.5) means that for any uAHsðGÞ;ðGðsÞðy1; Þ; uÞs ¼ uðy1Þ: ð0:6Þ

Therefore for any orthonormal basis fung of HsðRÞ; we haveGðsÞðy1; y2Þ ¼

Xn

unðy1Þunðy2Þ:

If we take

u0 ¼ 1; u2n�1ðyÞ ¼ffiffiffi2

pcos ny

ð1þ n2Þs=2; u2nðyÞ ¼

ffiffiffi2

psin ny

ð1þ n2Þs=2; nX1; ð0:7Þ

we obtain

GðsÞðy1; y2Þ ¼XnAZ

effiffiffiffiffi�1

pnðy1�y2Þ

ð1þ n2Þs : ð0:8Þ

Let fxn;a; nX1; a ¼ 1;y; dg be a sequence of independent standard real-valued

Brownian motion, defined on a probability space ðO;F;PÞ: Considerxðt; yÞ ¼

Xn;a

xn;aðtÞen;aðyÞ: ð0:9Þ

It is well known that for s412the above series converges uniformly with respect to

ðt; yÞA½0; 1� S1: We have for a; bAG;

Eð/xðt; y1Þ; aSG/xðs; y2Þ; bSGÞ ¼ t4sGðy1; y2Þ/a; bSG: ð0:10Þ

ARTICLE IN PRESSS. Fang / Journal of Functional Analysis 213 (2004) 440–465 441

t-xðt; Þ is a Brownian motion on CðS1;GÞ with ð ; Þs as covariance operator. For

yAS1; consider the following s.d.e.

dtgxðt; yÞ ¼ gxðt; yÞ 3 dtxðt; yÞ; gxð0; yÞ ¼ e: ð0:11Þ

It has been proved in [Ma] that ðt; yÞ-gxðt; yÞ has a continuous version that we

denote by the same notation. Then t-gxðt; Þ is a continuous process on LðGÞ :¼CðS1;GÞ: Denote by nðsÞt the law of x-gxðt; Þ on LðGÞ: The geometric stochasticanalysis on nðsÞt is well developed for s41

2: See [Ca,DL,Dr,Fa] for s ¼ 1 and [FG,In]

for s41=2: The case where s ¼ 12is critical but it is involved in Mathematical Physics.

On the based loop group LeðGÞ ¼ LðGÞ=G; there is a complex structure J which iscompatible with the metric H1=2 (see [Fr,PS]). However the H1=2 loops in G do not

form a smooth manifold. It is the limit case of HsðGÞ as sk1=2:The paper is organized as follows. In Section 1, we shall prove that the

expression of Ricci tensor obtained in [In] for s41=2 is valid for sX1=2 and

Ricsðz1; z2Þ converges to Ric1=2ðz1; z2Þ when s-1=2: In Section 2, we shall simplifythe Inahama’s expression and prove the boundedness of Rics for s41=2 by

estimating as well as possible the norm jjRicsjjop: The boundedness of Ric1=2 will be

established in Section 3. In Section 4, we shall discuss the uniform boundednessabout jjRicsjjop:

1. Ricci tensor in HsðGÞ for sX12

Let z1; z2ACNðS1;GÞ: Define the bracket by

½z1; z2�ðyÞ ¼ ½z1ðyÞ; z2ðyÞ�: ð1:1Þ

Following [Fr, p. 230], we define for s40;

rz1z2 ¼1

2f½z1; z2� þ ð1þ DÞ�s½z1; ð1þ DÞs

z2� þ ð1þ DÞ�s½z2; ð1þ DÞsz1�g: ð1:2Þ

We see that rz1z2ACNðS1;GÞ and satisfies

ðrz1z2; hÞs ¼1

2fð½z1; z2�; hÞs � ð½z1; h�; z2Þs � ð½z2; h�; z1Þsg: ð1:3Þ

Proposition 1.1 (Inahama [In]). Let z1; z2ACNðS1;GÞ; then

ðrz1z2ÞðyÞ ¼XkAZ

Xpþq¼k

½z1ðpÞ; z2ðqÞ�Tp;q

!effiffiffiffiffi�1

pky; ð1:4Þ

ARTICLE IN PRESSS. Fang / Journal of Functional Analysis 213 (2004) 440–465442

where

Tp;q ¼ 1

2

ð1þ ðp þ qÞ2Þs � ð1þ p2Þs þ ð1þ q2Þs

ð1þ ðp þ qÞ2Þs: ð1:5Þ

Proof. We have

½z1; z2� ¼XkAZ

Xpþq¼k

½z1ðpÞ; z2ðqÞ� !


pky;

ð1þ DÞ�1½z1; ð1þ DÞsz2� ¼

XkAZ

Xpþq¼k

½z1ðpÞ; z2ðqÞ�ð1þ q2Þs

ð1þ ðp þ qÞ2Þs


pky;

ð1þ DÞ�1½z2; ð1þ DÞsz1� ¼ �

XkAZ

Xpþq¼k

½z1ðpÞ; z2ðqÞ�ð1þ p2Þs

ð1þ ðp þ qÞ2Þs


pky:

Therefore by (1.2), we obtain the result (1.4). &

We shall compute the associated Ricci curvature. For z1; z2; z3ACNðS1;GÞ; define

Rðz1; z2Þz3 ¼ rz1rz2z3 �rz2rz1z3 �r½z1;z2�z3: ð1:6Þ

By Freed [Fr], the two-step trace of ðh1; h2Þ-ðRðz1; h1Þh2; z2Þs does exist if s414:

More precisely, the series

Ricsðz1; z2Þ :¼XnX0

Xd

a¼1ðRðz1; en;aÞen;a; z2Þs

exists for s41=4: For a; bAG; let

Kða; bÞ ¼ TraceðadðaÞ 3 adðbÞÞ:

When G is semi-simple, �K is an AdG invariant metric on G: For simplicity ofnotations, we shall take �K as the AdG invariant metric on G in the sequel.

Lemma 1.2. Let

AðsÞq ðcÞ ¼ �Tc;�qTq;c�q þ Tcþq;�q

ð1þ q2Þs for sX1=2: ð1:7Þ


Then there exists a constant C40 (independent of c) such that

sup1=2psp1

jAðsÞq ðcÞjpC

c2

q2for jqjX3c: ð1:8Þ

Proof. Set

NðsÞq ðcÞ ¼ ð1þ c2Þ2s þ ð1þ q2Þ2s � ð1þ ðc� qÞ2Þ2s � 2ð1þ c2Þsð1þ q2Þs

þ 2ð1þ c2Þsð1þ ðc� qÞ2Þs � 2ð1þ ðcþ qÞ2Þsð1þ ðc� qÞ2Þs

þ 2ð1þ q2Þsð1þ ðc� qÞ2Þs;

and

DðsÞq ðcÞ ¼ ð1þ q2Þsð1þ c2Þsð1þ ðc� qÞ2Þs:

Then by (1.5) and (1.7), we get AðsÞq ðcÞ ¼ 1

4

NðsÞq ðcÞ

DðsÞq ðcÞ

: Remark that there exists a constant

C40 (independent of gA½�2; 4�Þ such that

jð1þ xÞg � ð1þ gxÞjpCx2; jxjp12; gA½�2; 4�;

which will be written in the form

ð1þ xÞg ¼ 1þ gx þ Oðx2Þ; jxjp1

2; ð1:9Þ

where O is uniform with respect to gA½�2; 4�: Let c40 and jqjX3c: To fix the idea,consider q40: Then we have

ð1þ q2Þ2s ¼ q4s 1þ O1

q2

� �� :

To estimate the term ð1þ ðq � cÞ2Þs; remark that for qX3c; q � cX2cX2: Applying(1.9), we get

1þ 1

ðq � cÞ2

!2s

¼ 1þ 2s

ðq � cÞ2þ O

1

ðq � cÞ2

!

¼ 1þ O1

q2

� �; as q � cXq=2;

and

ðq � cÞ4s ¼ q4s 1� 4sc

qþ O

c2

q2

� �� :


Therefore

ð1þ ðq � cÞ2Þ2s ¼ q4s 1� 4sc

qþ O

c2

q2

� �� :

In the same way, we find

ð1þ ðq þ cÞ2Þs ¼ q2s 1þ 2sc

qþ O

c2

q2

� �� ;

ð1þ ðq � cÞ2Þsð1þ ðq þ cÞ2Þs ¼ q4s 1þ Oc2

q2

� �� ;

ð1þ q2Þsð1þ ðc� qÞ2Þs ¼ q4s 1� 2sc

qþ O

c2

q2

� �� :

A straightforward calculation yields

NsqðcÞ ¼ ð1þ c2Þ2s � 4scð1þ c2Þs

q2s�1 þ c2Oðq4s�2Þ:

For term DðsÞq ðcÞ; we have for qX3c; ð1þ ðq � cÞ2Þs

Xðq=2Þ2s: Then

DðsÞq ðcÞXð1þ c2Þs

q4s 1

2

� �2s

:

Therefore there exists a constant C40 such that

sup1=2psp1

NsqðcÞ

DsqðcÞ

��pC

c2

q2for qX3c: ð1:10Þ

For qp� 3c; changing q into �q; the term ðq � cÞ2 goes into ðq þ cÞ2 and ðq þ cÞ2

into ðq � cÞ2: So estimate (1.10) remains true for qp� 3c: The proof of (1.8) iscomplete. &

Lemma 1.3. There exists a constant C40 such that

sup1=2psp1

XqAZ

jAðsÞq ðcÞj

!pCc2: ð1:11Þ

Proof. Remark that for jqjp3c;

jNsqðcÞjpCð1þ c2Þ2s for some constant C40:


Then for sX1=2;

Xjqjo3c

NsqðcÞ

DsqðcÞ

��pCð1þ cÞs

Xjqjo3c

1ffiffiffiffiffiffiffiffiffiffiffiffiffi1þ q2

p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðq � cÞ2

q :

We have

Xcq¼0



q p2X½c=2�þ1q¼0



q p2ð½c=2� þ 2Þ½c=2� � 1

p4;

where ½c=2� denotes the integral part of c=2; and

X3cq¼cþ1



q p2:

It follows that there exists a constant C140 such that

Xjqjo3c

NsqðcÞ

DsqðcÞ

��pC1ð1þ c2Þs:

According to (1.8), we obtain (1.11). &

Theorem 1.4 (see Inahama [In]). Let z1; z2ACNðS1;GÞ: For sX1=2; we have

Ricsðz1; z2Þ ¼XcAZ

ð1þ c2ÞsXqAZ

AðsÞq ðcÞ

!/z1ðcÞ; %z2ðcÞS; ð1:12Þ

where ‘‘bar’’ denotes the complex conjugation and the inner product of G is extended by

complex linearity to its complexification GC:

Proof. The expression (1.12) comes from [In, (2.31)] with an opposite sign. In whatfollows, we shall give a proof valid for all case sX1=2: By (1.4), for zAHsðRÞ#G; wehave: rzz ¼ 0: Then Rðz1; zÞz ¼ �rzrz1z �r½z1;z�z: We have

rzrz1z ¼XcAZ

Xrþk¼c

Xpþq¼k

½zðrÞ; ½z1ðpÞ; zðqÞ��Tp;qTr;k


pcy;

r½z1;z�z ¼XcAZ

Xkþr¼c

Xpþq¼k

½½z1ðpÞ; zðqÞ�; zðrÞ�Tk;r


pcy:


Then

Rðz1; zÞz ¼XcAZ

Xrþk¼c

Xpþq¼k

½zðrÞ; ½z1ðpÞ; zðqÞ��ð�Tp;qTr;k þ Tk;rÞ !


pcy:

Under the hypothesis of z1; zACNðS1;GÞ; the above multiple series converge

absolutely. Let z2ACNðS1;GÞ: Consider en;a defined in (0.4) and (0.7), then

ðRðz1; en;aÞen;a; z2Þs

¼XcAZ

ð1þ c2ÞsX

pþqþr¼c

/½ea; ½z1ðpÞ; ea��; %z2ðcÞSunðrÞunðqÞð�Tp;qTr;pþq þ Tpþq;rÞ !

:

Therefore

Xd


¼Xc

ð1þ c2ÞsX

pþqþr¼c

/z1ðpÞ; %z2ðcÞSunðrÞunðqÞð�Tp;qTr;pþq þ Tpþq;rÞ:

Remark that

u2n�1ðrÞu2n�1ðqÞ þ u2nðrÞu2nðqÞ

is equal to 1ð1þn2Þs if r ¼ �q ¼ n or r ¼ �q ¼ �n; and equal to zero otherwise.

Therefore

X2N

n¼0

Xd


¼XN

n¼�N

Xc

ð1þ c2Þs

ð1þ n2Þs ðTc;�nTn;c�n � Tc�n;nÞ/z1ðcÞ; %z2ðcÞS:

Splitting the sum into two terms and changing the index n into �n in the secondterm, the above sum goes to

XN

n¼�N

Xc

ð1þ c2Þs

ð1þ n2Þs ðTc;�nTn;c�n � Tcþn;�nÞ/z1ðcÞ; %z2ðcÞS:

Therefore

X2N

n¼0

Xd


¼Xc

ð1þ c2ÞsXN

n¼�N

AðsÞn ðcÞ

!/z1ðcÞ; %z2ðcÞS: ð1:13Þ


Now by (1.11), for z1; z2ACNðS1;GÞ;XcAZ

ð1þ c2Þ2X

q

jAðsÞq ðcÞj

!j/z1ðcÞ; %z2ðcÞSjoþN:

Letting N-þN in (1.13), we get the result (1.12) by Lebesgue dominatedconvergence theorem. &

Theorem 1.5. For z1; z2ACNðS1;GÞ;

limsk1=2

Ricsðz1; z2Þ ¼ Ric1=2ðz1; z2Þ: ð1:14Þ

Proof. By (1.8), we see that the seriesP

q AðsÞq ðcÞ converges uniformly in sA½1

2; 1�:

Then

limsk1=2

Xq

AðsÞq ðcÞ ¼

Xq

Að1=2Þq ðcÞ:

Now for z1; z2ACNðS1;GÞ;XcAZ

ð1þ c2Þ2j/z1ðcÞ; %z2ðcÞSjoþN:

Again according to (1.11), we get

limsk1=2

Ricsðz1; z2Þ ¼XcAZ

ð1þ c2Þ1=2XqAZ

Að1=2ÞðcÞ !

/z1ðcÞ; %z2ðcÞS

¼Ric1=2ðz1; z2Þ: &

2. Boundedness of Rics for s41=2

In [In], it has been proved that for s41=2; there exists a constant Cs40 such that

jRicsðz1; z2ÞjpCsjz1jsjz2js; z1; z2AHsðGÞ: ð2:1Þ

It follows that there exists a bounded operator Rics :HsðGÞ-HsðGÞ such that

ðRicsz1; z2Þs ¼ Ricsðz1; z2Þ:

In what follows, we shall simplify expression (1.12) and estimate as well as possiblethe norm jjRicsjjop:


Proposition 2.1. Define

BðsÞq ðcÞ ¼ ð1þ c2Þs

ð1þ ðc� qÞ2Þsð1þ q2Þs; ð2:2Þ

DðsÞq ðcÞ ¼ ½ð1þ q2Þs � ð1þ ðcþ qÞ2Þs� � ½ð1þ ðq � cÞ2Þs � ð1þ qÞ2Þs�

ð1þ c2Þsð1þ q2Þs : ð2:3Þ

Then for s41=2;

4XqAZ

AðsÞq ðcÞ ¼

XqAZ

BðsÞq ðcÞ þ

XqAZ

DðsÞq ðcÞ: ð2:4Þ

Proof. The term NðsÞq ðcÞ defined in proof of Lemma 1.2 can be written in

the form:

NðsÞq ðcÞ ¼ 2ð1þ c2Þsð1þ ðc� qÞ2Þs � 2ð1þ c2Þsð1þ q2Þs

þ ð1þ c2Þ2s þ ½ð1þ q2Þ2s � ð1þ ðcþ qÞ2Þsð1þ ðc� qÞ2Þs�

þ ½2ð1þ q2Þs � ð1þ ðcþ qÞ2Þs � ð1þ ðc� qÞ2Þs�ð1þ ðc� qÞ2Þs:

Then

4AðsÞq ðcÞ ¼ 2

ð1þ q2Þs �2

ð1þ ðq � cÞ2Þsþ ð1þ c2Þs

ð1þ q2Þsð1þ ðq � cÞ2Þs

þ ð1þ q2Þ2s � ð1þ ðcþ qÞ2Þsð1þ ðc� qÞ2Þs

ð1þ c2Þsð1þ q2Þsð1þ ðq � cÞ2Þs

þ 2ð1þ q2Þs � ð1þ ðcþ qÞ2Þs � ð1þ ðc� qÞ2Þs

ð1þ c2Þsð1þ q2Þs :

For s41=2;P

qAZ1

ð1þq2Þs ¼P

qAZ1

ð1þðq�cÞ2Þs: Therefore in order to get (2.4), it is

sufficient to prove that

XqAZ

ð1þ q2Þ2s � ð1þ ðcþ qÞ2Þsð1þ ðc� qÞ2Þs

ð1þ c2Þsð1þ q2Þsð1þ ðq � cÞ2Þs¼ 0: ð2:5Þ


Remark firstly that the series in (2.5) converges absolutely if s41=2: In fact for

0osp1; using the inequality jxs � ysjpjx � yjs; we have

jð1þ q2Þ2s � ð1þ ðcþ qÞ2Þsð1þ ðc� qÞ2Þsj

pj2c2 þ c4 � 2c2q2js

Bð2c2Þsq2s; jqj-þN:

Now

Xkcþc

q¼kcþ1

ð1þ q2Þs

ð1þ ðq � cÞ2Þs� ð1þ ðcþ qÞ2Þs

ð1þ q2Þs

!

¼Xkc

q¼kc�cþ1

ð1þ ðcþ qÞ2Þs

ð1þ q2Þs �Xkcþc

q¼kcþ1

ð1þ ðcþ qÞ2Þs

ð1þ q2Þs :

Then

XN

k¼�N

Xkcþc

q¼kcþ1

ð1þ q2Þs

ð1þ ðq � cÞ2Þs� ð1þ ðcþ qÞ2Þs

ð1þ q2Þs

!" #

¼X�Nc

q¼�Nc�cþ1

ð1þ ðcþ qÞ2Þs

ð1þ q2Þs �XNcþc

q¼Ncþ1

ð1þ ðcþ qÞ2Þs

ð1þ q2Þs

which converges to c� c ¼ 0 as N-þN: We get therefore (2.5). &

Lemma 2.2. Define for s41=2;

as ¼XqAZ

1

ð1þ q2Þs: ð2:6Þ

Then we have the following estimate

supcAZ

XqAZ

ð1þ c2Þs

ð1þ ðc� qÞ2Þsð1þ q2Þspð1þ 4sÞas: ð2:7Þ

Proof. Let c40: Then

Xqo0

ð1þ c2Þs

ð1þ ðc� qÞ2Þsð1þ q2ÞspXqo0

1

ð1þ q2Þs;

XqXc

ð1þ c2Þs

ð1þ ðc� qÞ2Þsð1þ q2ÞspXqXc

1

ð1þ ðc� qÞ2Þs¼XqX0

1

ð1þ q2Þs:


We have

Xc�1q¼0

ð1þ c2Þs

ð1þ ðc� qÞ2Þsð1þ q2Þs

¼X½c=2�q¼0

ð1þ c2Þs

ð1þ ðc� qÞ2Þsð1þ q2Þsþ

Xc�1q¼½c=2�þ1

ð1þ c2Þs

ð1þ ðc� qÞ2Þsð1þ q2Þs:

It is easy to see that the first term on the right side is dominated by 22sP

qX01

ð1þq2Þs;

while the second term is dominated by 22sP

q401

ð1þq2Þs: Now combining these

estimates, we obtain (2.7). &

Lemma 2.3. Let 12osp1: We haveX

qAZ

jDðsÞq ðcÞjpð2þ 4sÞas: ð2:8Þ

Proof. Let c40: Remark that for 12osp1; the function f ðxÞ ¼ ð1þ xÞs is concave

and gðxÞ ¼ ð1þ x2Þs is convex. We have

2ð1þ c2Þspð1þ ðcþ qÞ2Þs þ ð1þ ðc� qÞ2Þsp2ð1þ c2 þ q2Þs: ð2:9Þ

Therefore for 0pqpc;

jð1þ ðcþ qÞ2Þs þ ð1þ ðc� qÞ2Þs � 2ð1þ q2Þsj

p2½ð1þ c2 þ q2Þs � ð1þ q2Þs�p2cs:

It follows that

Xcq¼0

jDðsÞq ðcÞjpas:

For q4c; remark that g00ðxÞ ¼ 2sð1þ x2Þs�2 þ 2sð2s � 1Þx2ð1þ x2Þs�1: Write down

ð1þ ðq þ cÞ2Þs � ð1þ q2Þs ¼ 2sg1ð1þ g21Þsc; g1A�q; q þ c½;

ð1þ q2Þs � ð1þ ðq � cÞ2Þs ¼ 2sg2ð1þ g22Þsc; g2A�q � c; q½:

Then there exists bA�g1; g2½C�q � c; q þ c½ such that

j½ð1þ ðq þ cÞ2Þs � ð1þ q2Þs� � ½ð1þ q2Þs � ð1þ ðq � cÞ2Þs�j

p4sc2ð1þ b2Þs�2 þ 4sð2s � 1Þc2ð1þ b2Þs�1 ð2:10Þ


which is smaller than 8sc2

ð1þðq�cÞ2Þ1�s: It follows that

Xq4c

jDðsÞq ðcÞjp 8s

Xq4c

c2

ð1þ c2Þsð1þ q2Þsð1þ ðq � cÞ2Þ1�s

¼ 8sXq40

c2

ð1þ c2Þsð1þ ðq þ cÞ2Þsð1þ q2Þ1�s:

Remarking that ð1þ ðq þ cÞ2ÞsXð1þ c2Þ1�sð1þ q2Þ2s�1; we get

Xq4c

jDðsÞq ðcÞjp8s

Xq40

1

ð1þ q2Þsp4sas: &

Proposition 2.4. For s41=2; we have

jjRicsjjoppð3þ 4s þ 4sÞas

4: ð2:11Þ

Proof. Using expression (1.12) and estimates (2.4), (2.7) and (2.8), we get

jðRicsz1; z2Þsjpð3þ 4s þ 4sÞas

4jz1jsjz2js: &

It has been proved in [In] (see also [FG]) that z-rzh is Hilbert–Schmidt operatorin HsðGÞ for s41=2 and jjrhjjHSpCsjhjs: This is the key point to develop the

geometric analysis on LðGÞ (see [Dr,Fa]). In this case, there exists a boundedoperator K :HsðGÞ-HsðGÞ such that

ðKz1; z2Þs ¼ �Xn;m

Xd

a;b¼1ðren;az1; em;bÞsðrem;bz2; en;aÞs: ð2:12Þ

The operator K involved naturally in [Fa]. For H1 metric, it is known that K is

identical to Ric1 by their explicit expression (see [FF]). In what follows, we shallprove that it is also true for s41=2:

Proposition 2.6. K ¼ Rics for s41=2:

Proof. By direct computations, we have for s41=2;

ðKz1; z2Þs ¼XcAZ

ð1þ c2ÞsXqAZ

Tq;cTqþc;�c

ð1þ c2Þs

!/z1ðcÞ; %z2ðcÞS: ð2:13Þ


Let q; cAZ; denote

gðc; qÞ ¼ ð1þ c2Þs � ð1þ q2Þs þ ð1þ ðc� qÞ2Þs: ð2:14Þ

Then we have

�Tc;�qTq;c�q þ Tcþq;�q

ð1þ q2Þs ¼ 1

4

�gðq; cÞgðc; qÞ þ 2gðc; cþ qÞð1þ ðc� qÞ2Þs

ð1þ q2Þsð1þ c2Þsð1þ ðc� qÞ2Þs:

and

Tq;cTqþc;�c

ð1þ c2Þs ¼ 1

4

gðq þ c; qÞ gðq; q þ cÞð1þ q2Þsð1þ c2Þsð1þ ðcþ qÞ2Þs

: ð2:15Þ

By changing q to �q in (2.15), we get

XqAZ

Tq;cTqþc;�c

ð1þ c2Þs ¼ 1

4

XqAZ

gðc; qÞgðq; q � cÞð1þ q2Þsð1þ c2Þsð1þ ðc� qÞ2Þs

:

By (2.14), we see that

�gðq; cÞ � gðq; q � cÞ ¼ �2ð1þ q2Þs;

and

� gðc; qÞð1þ q2Þs þ gðc; cþ qÞð1þ ðc� qÞ2Þs

¼ ð1þ c2Þs½ð1þ ðc� qÞ2Þs � ð1þ q2Þs� þ ½ð1þ q2Þ2s � ð1þ ðc� qÞ2Þsð1þ ðcþ qÞ2Þs�:

It follows that

XqAZ

�Tc;�qTq;c�q þ Tcþq;�q

ð1þ q2Þs �XqAZ

Tq;c Tqþc;�c

ð1þ c2Þs

¼XqAZ

�2gðc; qÞð1þ q2Þs þ 2gðc; cþ qÞð1þ ðc� qÞ2Þs

4ð1þ q2Þsð1þ c2Þsð1þ ðq � cÞ2Þs

¼ 1

2

XqAZ

ð1þ c2Þs½ð1þ ðc� qÞ2Þs � ð1þ q2Þs�ð1þ q2Þsð1þ c2Þsð1þ ðq � cÞ2Þs

þ 1

2

XqAZ

ð1þ q2Þ2s � ð1þ ðc� qÞ2Þsð1þ ðcþ qÞ2Þs

ð1þ q2Þsð1þ c2Þsð1þ ðq � cÞ2Þs

¼ I1ðcÞ þ I2ðcÞ:


By taking k ¼ c� q in the summation of I1ðcÞ; we get

I1ðcÞ ¼ �I1ðcÞ:

Therefore I1ðcÞ ¼ 0: I2ðcÞ ¼ 0 follows from (2.5). &

3. Boundedness of Ric1=2

The main purpose of this section is to prove

Theorem 3.1. Let z1; z2ACNðS1;GÞ; then

jRic1=2ðz1; z2Þjpð4pÞjz1j1=2jz2j1=2: ð3:1Þ

Proof. Denote AqðcÞ ¼ 4Að1=2Þq ðcÞ: Then

AqðcÞ ¼1þ 2qc� 2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1þ c2Þð1þ q2Þ

qþ 2½

ffiffiffiffiffiffiffiffiffiffiffiffiffi1þ c2

pþ

ffiffiffiffiffiffiffiffiffiffiffiffiffi1þ q2

p�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðcþ qÞ2

q�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðc� qÞ2

qffiffiffiffiffiffiffiffiffiffiffiffiffi1þ c2

p ffiffiffiffiffiffiffiffiffiffiffiffiffi1þ q2

p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðc� qÞ2

q :

By (1.12), we have

Ric1=2ðz1; z2Þ ¼1

4

XcAZ


p XqAZ

AqðcÞ !

/z1ðcÞ; %z2ðcÞS: ð3:2Þ

Using the following lemma, we get the result. &

Lemma 3.2. We have

supc

XqAZ

jAqðcÞjp16p: ð3:3Þ

Proof. By symmetry, it is sufficient to prove that supcX0

PqAZ jAqðcÞjp16p:

Consider firstly the termP

qX0 jAqðcÞj: Define

A1qðcÞ ¼

qc�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1þ c2Þð1þ q2Þ



p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðc� qÞ2

q ;

A2qðcÞ ¼


pþ


p�




p :


We have

qcffiffiffiffiffiffiffiffiffiffiffiffiffi1þ c2


p � 1 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffi1þ q2

p � 1

!� 1� cffiffiffiffiffiffiffiffiffiffiffiffiffi

1þ c2p

!qffiffiffiffiffiffiffiffiffiffiffiffiffi

1þ q2p

¼� 1ffiffiffiffiffiffiffiffiffiffiffiffiffi1þ q2

pðq þ


pÞ� 1ffiffiffiffiffiffiffiffiffiffiffiffiffi

1þ c2p

ðcþffiffiffiffiffiffiffiffiffiffiffiffiffi1þ c2

pÞ

qffiffiffiffiffiffiffiffiffiffiffiffiffi1þ q2

p :

Let

B1qðcÞ ¼ � 1ffiffiffiffiffiffiffiffiffiffiffiffiffi

1þ q2p

ðq þffiffiffiffiffiffiffiffiffiffiffiffiffi1þ q2

pÞ

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðc� qÞ2

q ;

B2qðcÞ ¼

1ffiffiffiffiffiffiffiffiffiffiffiffiffi1þ c2

pðcþ


pÞ


p 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðq þ cÞ2

q � 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðq � cÞ2

q0B@

1CA;

B3qðcÞ ¼ � 1ffiffiffiffiffiffiffiffiffiffiffiffiffi

1þ c2p

ðcþffiffiffiffiffiffiffiffiffiffiffiffiffi1þ c2

pÞ


p 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðq þ cÞ2

q :

Then

A1qðcÞ ¼ B1

qðcÞ þ B2qðcÞ þ B3

qðcÞ: ð3:4Þ

Clearly

jB1qðcÞjp

1

1þ q2and

Xcq¼0

jB2qðcÞjp1: ð3:5Þ

We have

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðq þ cÞ2

q � 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðq � cÞ2

q

¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðq � cÞ2

q�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðq þ cÞ2

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðq þ cÞ2

q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðq � cÞ2

q¼ �4qcffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1þ ðq þ cÞ2q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1þ ðq � cÞ2q

ðffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðq � cÞ2

qþ


qÞ:


It follows that

Xq4c

jB2qðcÞjp

XqX1

1

1þ q2oþN: ð3:6Þ

For the estimate of A2qðcÞ; we have


p�




p ¼ �c2 � 2qcffiffiffiffiffiffiffiffiffiffiffiffiffi1þ c2


pðffiffiffiffiffiffiffiffiffiffiffiffiffi1þ q2

pþ


qÞ:

Let

C1qðcÞ ¼


p � cþ 2qffiffiffiffiffiffiffiffiffiffiffiffiffi1þ q2


pþ


q ;

C2qðcÞ ¼

cffiffiffiffiffiffiffiffiffiffiffiffiffi1þ c2

pðcþ


pÞ



pþ


qÞ;

C3qðcÞ ¼

1ffiffiffiffiffiffiffiffiffiffiffiffiffi1þ c2

pðcþ


pÞ


p 2ffiffiffiffiffiffiffiffiffiffiffiffiffi1þ q2

pþ


q :

Then

A2qðcÞ ¼ C1

qðcÞ þ C2qðcÞ þ C3

qðcÞ: ð3:7Þ

Clearly

jC2qðcÞjp

1

1þ q2: ð3:8Þ

We have

C1qðcÞ ¼


pffiffiffiffiffiffiffiffiffiffiffiffiffi1þ q2

pþ


q� ðq þ cþ qÞffiffiffiffiffiffiffiffiffiffiffiffiffi

1þ q2p

þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðcþ qÞ2

q

¼ 2ffiffiffiffiffiffiffiffiffiffiffiffiffi1þ q2

p 1þffiffiffiffiffiffiffiffiffiffiffiffiffi1þ q2

p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðcþ qÞ2

q� qðcþ qÞ

ðffiffiffiffiffiffiffiffiffiffiffiffiffi1þ q2

pþ


qÞð


pþ


qþ cþ 2qÞ


and



q� qðcþ qÞ ¼ 1þ q2 þ ðcþ qÞ2ffiffiffiffiffiffiffiffiffiffiffiffiffi

1þ q2p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1þ ðcþ qÞ2q

þ qðcþ qÞ:

It follows that

jC1qðcÞjp

4

1þ q2: ð3:9Þ

We have


pþ


q � 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðq þ cÞ2

q

¼


q�


pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðq þ cÞ2

qðffiffiffiffiffiffiffiffiffiffiffiffiffi1þ q2

pþ


qÞ

¼ c2 þ 2qcffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðq þ cÞ2

qðffiffiffiffiffiffiffiffiffiffiffiffiffi1þ q2

pþ


qÞ2:

It follows that

jB3qðcÞ þ C3

qðcÞjp1

1þ q2: ð3:10Þ

Combining (3.4)–(3.10) and remarking thatP

qX01ffiffiffiffiffiffiffiffi

1þc2p ffiffiffiffiffiffiffiffi

1þq2p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1þðc�qÞ2p is dominated

by 1þP

qX11

1þq2; we obtain that

supcX0

XqX0

jAqðcÞjp9XqX0

1

1þ q2

!:

We shall estimate the termP

qo0 jAqðcÞj: We have for q40

A�qðcÞ ¼1� 2qc� 2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1þ c2Þð1þ q2Þ

qþ 2½


pþ


p�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðc� qÞ2

q�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðcþ qÞ2

q�ffiffiffiffiffiffiffiffiffiffiffiffiffi

1þ c2p ffiffiffiffiffiffiffiffiffiffiffiffiffi


1þ ðcþ qÞ2q :


Let

D1qðcÞ ¼

�2qcþ ðffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðcþ qÞ2

q�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðc� qÞ2

qÞffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðcþ qÞ2




q ;

D2qðcÞ ¼

qc�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1þ c2Þð1þ q2Þ




q þ


pþ


p�




p :

We have

D1qðcÞ ¼

1þ ðc2 þ q2 �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðc� qÞ2

q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðcþ qÞ2

qÞffiffiffiffiffiffiffiffiffiffiffiffiffi

1þ c2p ffiffiffiffiffiffiffiffiffiffiffiffiffi


1þ ðcþ qÞ2q :

As

c2 þ q2 �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðc� qÞ2


q¼ 4c2q2 � 1� 2c2 � 2q2

ðc2 þ q2Þ þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðc� qÞ2


q ;

we get

jD1qðcÞjp

1

1þ q2þ 4c

q2 þ c2: ð3:11Þ

According to the fact

XþN

q¼1

c

q2 þ c2pZ þN

0

c

t2 þ c2dt ¼ p

2;

we get

supcX0

XqX1

jD1qðcÞjp2pþ

XqX1

1

1þ q2: ð3:12Þ

For the estimate of D2qðcÞ; we have

D2qðcÞ ¼ � 1ffiffiffiffiffiffiffiffiffiffiffiffiffi

1þ q2p

ðq þffiffiffiffiffiffiffiffiffiffiffiffiffi1þ q2

pÞ

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ðcþ qÞ2

q þ B3qðcÞ þ A2

qðcÞ:

It follows that

jD2qðcÞjp

1

1þ q2þ jB3

qðcÞ þ C3qðcÞj þ jC1

qðcÞj þ jC2qðcÞj ð3:13Þ


which is smaller than 71þq2

; according to (3.10), (3.9) and (3.8). Combining (3.12) and

(3.13), we get

supcX0

Xqo0

jAqðcÞjp2pþ 9XqX1

1

1þ q2:

Finally, we get

XqAZ

jAqðcÞjp2pþ 9þ 18XqX1

1

1þ q2p16p: &

4. Further discussions on jjRicsjjop

The natural question is now whether

sup1=2psp1

jjRicsjjopoþN: ð4:1Þ

In what follows, we shall give a positive response for restriction on subspaces

generated by any finite subdivision. Let P ¼ fy1o?oyNg be a subdivision of S1:

Denote by GðsÞ ¼ ðGðsÞðyi; yjÞÞ1pi; jpN which is positive definite matrix. Let QðsÞ ¼ðQðsÞ

ij Þ1pi; jpN be the inverse of GðsÞ: Consider on GP ¼ G ? G the metric

jaj2P ¼XN

i; j¼1Q

ðsÞij /ai; ajSG; a ¼ ða1;y; aNÞAGP: ð4:2Þ

To a ¼ ða1;y; aNÞAGP; we associate haAHsðGÞ by

haðyÞ ¼XN

i; j¼1GðsÞðyi; yÞQðsÞ

ij aj: ð4:3Þ

By (0.6), we have jhaj2s ¼ jaj2P:

Proposition 4.1. We have for s412;

ðRics ha; haÞs ¼1

4

Xi; j;k;m

XcAZ

LðsÞc

e�ffiffiffiffiffi�1

pðyi�ykÞ

ð1þ c2Þs

!Q

ðsÞij Q

ðsÞkm/aj; amSG ð4:4Þ


where

LðsÞc ¼

XcAZ

BðsÞq ðcÞ þ

XcAZ

DðsÞq ðcÞ ðsee ð2:4ÞÞ:

Proof. Using (0.8), we have

haðcÞ ¼XN

i; j¼1

e�ffiffiffiffiffi�1

pcyi

ð1þ c2Þs QðsÞij aj:

Now applying (1.12) and (2.4), we get the result. &

Theorem 4.2. For P given, there exists a constant CP40 (independent of s) such that

jðRics ha; haÞsjpCPjhaj2s ; sA1

2; 1

� �: ð4:5Þ

To prove (4.5), we need some preparations.

Lemma 4.3. Let EðsÞ1 ðcÞ ¼

Pjqjpjcj D

ðsÞq ðcÞ: Then

XcAZ

EðsÞ1 ðcÞ

ð1þ c2Þs þXcAZ

1

ð1þ c2Þs

XqAZ

BðsÞq ðcÞ

��p2as þ 6p: ð4:6Þ

Proof. For cX1; we have

EðsÞ1 ¼ �2 1� 1

ð1þ c2Þs

� �þ 2

Xcq¼1

DðsÞq ðcÞ: ð4:7Þ

By (2.9), for 1pqpc; we have

2ð1þ c2Þs � 2ð1þ q2Þspð1þ ðcþ qÞ2Þs þ ð1þ ðc� qÞ2Þs � 2ð1þ q2Þsp2c2s:

Then

Xcq¼1

DðsÞq ðcÞ þ 2

Xcq¼1

1

ð1þ q2Þs

��p 2c

ð1þ c2Þsp2:


According to (4.7), we get

XcX1

EðsÞ1 ðcÞ

ð1þ c2Þs þ 4XNc¼1

Xcq¼1

1

ð1þ q2Þsð1þ c2Þs

��

p6XcX1

1

ð1þ c2Þsp3p: ð4:8Þ

Remark that a2s ¼ 4P

q;cX11

ð1þq2Þsð1þc2Þs þ 4P

qX11

ð1þq2Þs þ 1: By (4.8), we get

XcAZ

EðsÞ1 ðcÞ

ð1þ c2Þs þ a2s

��p2as þ 6p:

Now remarking that

XcAZ

1

ð1þ c2Þs

XqAZ

BðsÞq ðcÞ ¼

Xc;qAZ

1

ð1þ c2Þsð1þ q2Þs;

we get (4.6). &

Lemma 4.4. Let EðsÞ2 ðcÞ ¼

Pjqj4jcj D

ðsÞq ðcÞ: Then for 1=2psp1;

jEðsÞ2 ðcÞjp8 2þ

XqX1

1

ð1þ q2Þ3=2

!p8p: ð4:9Þ

Proof. Let c40: By (2.10), we get

Xq4c

jDðsÞq ðcÞjp 4s

Xq4c

c2


þ 4sð2s � 1ÞXq4c

c2


p 4sXqX1

1

ð1þ q2Þ1þsþ 4sð2s � 1Þ

XqX1

1

ð1þ q2Þs:

Remarking that

XqX1

1

ð1þ q2Þsp1þZ

N

1

dx

x2s¼ 2s

2s � 1;

we get (4.9). &


Lemma 4.5. Let gðsÞc ¼Pc

q¼1 DðsÞq ðcÞ; then

XcX1

jgðsÞcþ1 � gðsÞc jp4sþ1as þ 5p: ð4:10Þ

Proof. Write down

gðsÞcþ1 � gðsÞc ¼ �Xcþ1q¼1

ð1þ ðq þ cþ 1Þ2Þs þ ð1þ ðcþ 1� qÞ2Þs � 2ð1þ q2Þs

ðð1þ ðcþ 1Þ2Þsð1þ ðqÞ2Þs

þXcq¼1

ð1þ ðq þ cÞ2Þs þ ð1þ ðc� qÞ2Þs � 2ð1þ q2Þs

ðð1þ ðcÞ2Þsð1þ ðqÞ2Þs

¼ � ð1þ ð2ðcþ 1ÞÞ2Þs þ 1

ð1þ ðcþ 1Þ2Þ2sþ 2

cþ 1

ð1þ ðcþ 1Þ2Þs� c

ð1þ c2Þs

!

þXcq¼1

1þ ðq þ cÞ2Þs

ð1þ c2Þs � 1þ ðq þ cþ 1Þ2Þs

ð1þ ðcþ 1Þ2Þs

!1

ð1þ q2Þs

þXc�1q¼1

1þ ðc� qÞ2Þs

ð1þ c2Þs � 1þ ðcþ 1� qÞ2Þs

ð1þ ðcþ 1Þ2Þs

!1

ð1þ q2Þs

þ 1

ð1þ c2Þs �2s

ð1þ ðcþ 1Þ2Þs

!1

ð1þ c2Þs

¼ I1ðcÞ þ I2ðcÞ þ I3ðcÞ þ I4ðcÞ þ I5ðcÞ:

We have

(i)XcX1

jI1ðcÞjp4sXcX1

1

ð1þ c2Þs þXcX1

1

ð1þ c2Þ2s:

(ii)XcX1

jI5ðcÞjpXcX1

1þ 2s

ð1þ c2Þ2s:

For estimatingP

cX1 jI2ðcÞj; we remark that the function x- xð1þx2Þs is increasing

over ½1; 1ffiffiffiffiffiffiffiffi2s�1

p �; decreasing over ½ 1ffiffiffiffiffiffiffiffi2s�1

p ;N½: Let Ns ¼ ½ 1ffiffiffiffiffiffiffiffi2s�1

p �: Then

(iii)XcX1

jI2ðcÞjp2Ns

ð1þ N2s Þ

s þNs þ 1

ð1þ ðNs þ 1Þ2Þs

!p4:


ForP

cX1 jI3ðcÞj; we remark that the function x-ð1þðqþxÞ2Þs

ð1þx2Þs is decreasing over

½q;þN½: Therefore

jI3ðcÞj ¼Xcq¼1

1þ ðq þ cÞ2Þs

ð1þ c2Þsð1þ q2Þs �Xcq¼1

1þ ðq þ cþ 1Þ2Þs

ð1þ ðcþ 1Þ2Þsð1þ q2Þs:

It follows that

(iv)XcX1

jI3ðcÞjp2XcX1

ð1þ 4c2Þs

ð1þ c2Þ2sp2 4s

XcX1

1

ð1þ c2Þs:

In the same way, we get

(v)XcX2

jI4ðcÞjp2sXcX1

1

ð1þ c2Þ2sþXcX1

1

ð1þ c2Þs:

Combining (i)–(v), we get the result. &

Proof of Theorem 4.2. For iak; we have

XcAZ

XqAZ

BðsÞq ðcÞ

!e�ffiffiffiffiffi�1

pcðyi�ykÞ

ð1þ c2Þs ¼ ðGðsÞðyi; ykÞÞ2: ð4:11Þ

According to (4.7),

XcAZ

EðsÞ1 ðcÞ e�

ffiffiffiffiffi�1

pcðyi�ykÞ

ð1þ c2Þs ¼ 2XcX1

EðsÞ1 ðcÞ cosðcðyi � ykÞÞ

ð1þ c2Þs

¼ � 4XcX1

1� 1

ð1þ c2Þs

� �cosðcðyi � ykÞÞ

ð1þ c2Þs þ 4XcX1

gccosðcðyi � ykÞÞ

ð1þ c2Þs :

Let GðsÞN ¼

PNc¼1

cosðcðyi�ykÞÞð1þc2Þs ; Vcðyi; ykÞ ¼

PNc¼1 cosðcðyi � ykÞÞ and

aP ¼ supN;i;k

jVNðyi; ykÞj:

By Abel transformation, we get

supNX1

jGðsÞN jpjaPj

2s:


Again by Abel transformation, we obtain

XcX1

gccosðcðyi � ykÞÞ

ð1þ c2Þs

� ��p

XcX1

jgc � gcþ1j !

supN

jGðsÞN j

� �:

Now using (4.9) and (4.10), we get

XcAZ

XqAZ

DðsÞq ðcÞ

!e�ffiffiffiffiffi�1

pcðyi�ykÞ

ð1þ c2Þs

��

pð4þ 8pÞas þ 4ð2sas þ 5pÞaP: ð4:12Þ

Finally combining (4.6), (4.9), (4.11) and (4.12) and by (4.4), there exists a constantAP40 (independent of sA½1=2; 1�) such that

jðRics ha; haÞsjpAP

as

Xi; j;k;m

jasQðsÞij j jasQ

ðsÞk;mj j/aj; amSGj:

As s-1=2; asQðsÞij -dij : Let

BP ¼ supij

sup1=2osp1

jasQðsÞij joþN:

Then

jðRics ha; haÞsjpAPB2

P

as

N3XN

j¼1jaj j2G

!:

Let lPasbe the smallest eigenvalue of QðsÞ; then

jðRics ha; haÞsjpAPB2

PN3

lP

XN

i; j¼1Q

ðsÞij /ai; ajSG

!¼ APB2

PN3

lPjhaj2s : &

Remark 4.6. Let HPs ðGÞ ¼ fha; aAGPg and pP : HsðGÞ-HP

s ðGÞ be the orthogo-nal projection, then pP Rics pP is different of Ricci tensor Rics

P on HPs ðGÞ defined

by the induced metric. For this last one, we know that RicsPB

as

4Identity as s-1

2:

References

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California, San Diego, 1997.

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loop groups, Ann. Probab. 27 (1999) 664–683.

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391–407.

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Metrics H andbehavioursas sk1 2on loopgroups · 2017-02-08 ·...

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